Q.When the momentum of body is increased by three times, its K.E becomes .................

Showing the 12 most recent of 28 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the momentum of a body is increased by 100%, then the percentage increase in the kinetic energy is

   (a) 150%

   (b) 200%

   (c) 300%

   (d) None of these

   ›Reveal solutionSolution

   KE = p^2 / (2m), so KE scales with the SQUARE of momentum; doubling p means KE becomes 4 times as large, i.e. a 300% increase over the original.

   Kinetic energy in terms of momentum: KE = p^2 / (2m)

   Let the initial momentum be p, so initial KE = p^2/(2m).

   Momentum increased by 100% means the new momentum is p' = 2p.

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 markMCQ
   Q.When the momentum of body is increased by three times, its K.E. becomes:

   (a) Twice

   (b) Half

   (c) Four times

   (d) Nine times

   ›Reveal solutionSolution

   $KE = p^2/2m$, so $KE \propto p^2$ — tripling $p$ makes $KE$ nine times as large.

   Momentum is $p = mv$ and kinetic energy is $KE = \tfrac12 mv^2$. Writing $KE$ in terms of $p$: $KE = \tfrac12 mv^2 = \dfrac{(mv)^2}{2m} = \dfrac{p^2}{2m}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the linear momentum of the object is increased by 0.1% then the kinetic energy is increased by:

   (a) 0.4%

   (b) 0.1%

   (c) 0.01%

   (d) 0.2%

   ›Reveal solutionSolution

   Since KE = p^2/2m depends on the square of momentum, a small percentage change in p produces twice that percentage change in KE: 2 x 0.1% = 0.2%.

   Kinetic energy can be written in terms of linear momentum p = mv as

   KE = (1/2)mv^2 = p^2/(2m)

   Taking logarithms to relate fractional changes (mass m is constant):

   ln(KE) = 2 ln p - ln(2m)

   Differentiating:

   delta(KE)/KE = 2(deltap/p)

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Kinetic energy of a body of mass 2 kg is 36 joule. Its momentum in kg m/s will be

   (a) 6

   (b) 12

   (c) 24

   (d) 36

   ›Reveal solutionSolution

   p = sqrt(2 m KE) = sqrt(144) = 12 kg m/s. Answer (B).

   Kinetic energy in terms of momentum: KE = p^2/(2m), so p = sqrt(2 m KE).

   Given m = 2 kg and KE = 36 J:

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2026Set ANNUAL1 markMCQ
   Q.For constant momentum (p), the correct graph between kinetic energy (E) and mass (m) will be

   (a) decreasing hyperbola-like E-m curve

   (b) straight line increasing with mass

   (c) upward-increasing E-m curve

   (d) none of these

   ›Reveal solutionSolution

   With p constant, E = p^2/(2m) is proportional to 1/m -> a hyperbola-like decreasing curve. Answer (A).

   Kinetic energy can be written as E = p^2/(2m).

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Kinetic energies of two moving bodies of masses 1 gm and 4 gm are equal. The ratio of their linear momentum is (A) 4 : 1 (B) sqrt(2) : 1 (C) 1 : 2 (D) 1 : 16

   ›Reveal solutionSolution

   With equal kinetic energies, the ratio of momenta of the two bodies is 1:2.

   Kinetic energy in terms of momentum: $E_k = \dfrac{p^2}{2m} \Rightarrow p = \sqrt{2mE_k}$.

   Since both bodies have the same kinetic energy $E_k$:

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Two bodies of masses 2 kg and 32 kg have equal kinetic energy. What is the ratio of their momentum?

   (a) 1 : 1

   (b) 1 : 2

   (c) 1 : 4

   (d) 2 : 1

   ›Reveal solutionSolution

   Since KE = p^2/2m, momentum for a given kinetic energy is p = sqrt(2mKE); with equal KE, the momentum ratio reduces to sqrt(m1/m2).

   Kinetic energy: KE = p^2 / (2m), so p = sqrt(2m * KE).

   Both bodies have the same KE (call it E), so: …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set ANNUAL1 markMCQ
   Q.If the linear momentum of a body increases by 20%, what will be the percentage increase in K.E.?

   (a) 40%

   (b) 36%

   (c) 44%

   (d) 20%

   ›Reveal solutionSolution

   Since $KE = p^2/(2m)$, kinetic energy grows with the square of momentum — so a 20% rise in momentum produces a 44% rise in kinetic energy, not 20% or 40%.

   Kinetic energy expressed in terms of momentum $p$ and mass $m$:

   $KE = p^2/(2m)$

   If momentum increases by 20%, the new momentum is $p' = 1.2p$.

   New kinetic energy:

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The kinetic energy of a body of mass 1 kg is 1 joule. Its velocity will be (A) 0.45 ms^-1 (B) 1 ms^-1 (C) 1.4 ms^-1 (D) 4.4 ms^-1

   ›Reveal solutionSolution

   $v=\sqrt{2KE/m}\approx1.4\,\text{m/s}$.

   Kinetic energy: $KE=\tfrac12mv^2$. Rearranging, $v=\sqrt{\dfrac{2\,KE}{m}}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2024Set ANNUAL1 markMCQ
    Q.The linear momentum of a body of mass m is p. Then its kinetic energy will be (A) mp (B) mp^2 (C) p^2/m (D) p^2/2m

    ›Reveal solutionSolution

    In terms of momentum, $KE=p^2/2m$.

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The momentum of any body is increased by 20%. Then its kinetic energy will increase by (A) 20% (B) 66% (C) 44% (D) 88%

    ›Reveal solutionSolution

    A 20% rise in momentum raises kinetic energy by 44%.

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2024Set ANNUAL1 markMCQ
    Q.What is the kinetic energy of a 2 kg object moving at a velocity of 3 m/s ?

    (a) 6 J

    (b) 9 J

    (c) 18 J

    (d) 27 J

    ›Reveal solutionSolution

    Kinetic energy of a moving object is KE = ½mv²; here it works out to 9 J.

    The kinetic energy of an object of mass m moving with speed v is:

    $KE = \frac{1}{2}mv^2$

    Substituting m = 2 kg, v = 3 m/s:

    …

    🔒 Start your 14-day free trial to see the full solution →